In each question, two equations numbered I and II have been given. You have to solve both the equations and mark the appropriate option.
I.`x^2 +6x+9=0`
II. `y^2-y-20=0`

Options

A. `x lt y`
B. no relatioship can be established
C. `x le y`
D. `x ge y`

Answer

To solve the given equations step by step, let's break it down:

### Step 1: Solve the first equation \( I: x^2 + 6x + 9 = 0 \)

1. **Identify the equation**: The equation is a quadratic in the standard form \( ax^2 + bx + c = 0 \).
2. **Factor the equation**: We can factor the equation as follows:
\[ x^2 + 6x + 9 = (x + 3)(x + 3) = (x + 3)^2 = 0 \]
3. **Set the factors to zero**:
\[ (x + 3)^2 = 0 \]
4. **Solve for \( x \)**:
\[ x + 3 = 0 \implies x = -3 \]

### Step 2: Solve the second equation \( II: y^2 - y - 20 = 0 \)

1. **Identify the equation**: This is also a quadratic equation.
2. **Factor the equation**: We need to find two numbers that multiply to \(-20\) and add to \(-1\). These numbers are \(-5\) and \(4\):
\[ y^2 - 5y + 4y - 20 = (y - 5)(y + 4) = 0 \]
3. **Set the factors to zero**:
\[ (y - 5)(y + 4) = 0 \]
4. **Solve for \( y \)**:
\[ y - 5 = 0 \implies y = 5 \]
\[ y + 4 = 0 \implies y = -4 \]

### Step 3: Analyze the relationship between \( x \) and \( y \)

1. **Values obtained**: From the first equation, we have \( x = -3 \). From the second equation, we have \( y = 5 \) and \( y = -4 \).
2. **Compare the values**:
- \( x = -3 \)
- \( y = 5 \) and \( y = -4 \)

3. **Determine the relationship**:
- Comparing \( x \) and \( y \):
- \( x = -3 \) is less than \( y = 5 \)
- \( x = -3 \) is greater than \( y = -4 \)

Since we have two different values for \( y \) and one value for \( x \), we cannot establish a consistent relationship between \( x \) and \( y \) that holds true for both values of \( y \).

### Conclusion

The correct option is that there is no definitive relationship between \( x \) and \( y \).